3/ " Any 2 vertices connected by a path of length 3 have degree at most 2"

$\Longrightarrow$ Is this statement saying something about the "diagonal" ? Because I'm thinking about a picture of 4 vertices $A$, $B$, $C$, $D$ where $A \longrightarrow B$ and $B \longrightarrow C$ and $C \longrightarrow D$. So $A$ and $D$ have degree at most 2 when I connect $C$ with $A$ and $B$ with $D$.

1 Answer
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(1) No, it means that for any two distinct vertices $a$ and $b$ there are at least two different paths from $a$ to $b$, each of length $4$. If they’re disjoint, that means that the graph contains a circuit of length $8$ with $a$ and $b$ as far apart on the circuit as possible. HINT: Think of an octagon.

(2) This won’t quite work. The vertices of degree $2$ are $A,B$, and $C$, and the vertex of degree $3$ is $E$. $B$ and $C$ are connected to $E$ by paths of length $3$ (and also by paths of length $1$), but $A$ is not: both paths from $A$ to $E$ are of length $2$. Your graph can be drawn as a square with a ‘sticker’ at one corner ($D$) leading to $E$; what if you add another sticker like that at $A$, going to a new vertex $F$?

(3) Your graph is find before you add edges between $A$ and $C$ and between $B$ and $D$. The graph looks like this:

A --- B --- C --- D

The only vertices connected by a path of length $3$ are $A$ and $D$, and sure enough, they have degree at most $2$. ‘At most $2$’ is the same as $\le 2$. In fact, adding those two extra edges ruins the example, because then you have the path $BACD$ of length $3$ from $B$ to $D$, and yet $\deg B=3\not\le 2$.

Thank you for your response, but may I ask for your clarification on (2)? My thought is that if I have a vertex, say A, of degree 2, then there are exactly 2 edges incident to A. And if I have a vertex, say C, of degree 3, then there are exactly 3 edges incident to C. But how should I connect A and C ? Can you give an example of a picture which will do the job? I'm thinking about A --> B --> C --> D --> E --> A, but I feel something isn't quite right. Thank you
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CecileNov 11 '12 at 23:58

@Cecile: No, a circuit of five vertices won’t do the job. I’m not sure how that relates to what I suggested for (2), though, which was to take the graph that you originally had for (2) and add one more vertex connected to $A$.
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Brian M. ScottNov 12 '12 at 16:33